Atiyah-Bott-Shapiro generalization to $U(n) \to ({Spin(2n) \times U(1)})/{\mathbf{Z}/4}$ for $n=2k+1$

Question

Atiyah, Bott, and Shapiro paper on Clifford Modules around page 10 shows two facts.

1 - There is a lift $U(n) \to Spin^c(2n)$ from $U(n) \to SO(2n)\times U(1)$. Also an embedding (injective group homomorphism) $ U(n) \subset Spin^c(2n)$:

ABS show that a homomorphism $U(n) \to SO(2n)\times U(1)$ lifts to $Spin^c(2n)$ and give an explicit description of the lifting in terms of matrices.

Here is the homomorphism they wish to lift:

$l: U(n) \to SO(2n)\times U(1)$ given by $ T \mapsto j(T) \times \det(T)$. (Here $j: U(n) \to SO(2n)$).

Here is their lift $\tilde{l}: U(n) \to Spin^c(2n)$ :

Let $T \in U(n)$ be expressed relative to an orthonormal basis $f_1, \ldots, f_n$ of $\mathbb{C}^n$ by a diagonal matrix with diagonal entries $e^{it_1}, e^{it_2} , \ldots e^{it_n}$. Let $e_1,\ldots,e_{2n}$ be the corresponding basis of $\mathbb{R}^{2n}$, so that $e_{2j-1} = f_j$ and $e_{2j} = i f_j$. Then the corresponding element of $Spin^c(2n)$ is $$ \tilde{l}(T) = \prod_{j=1}^n \left( \cos (t_j/2) + \sin (t_j/2) e_{2j-1}e_{2j} \right) \times \exp( i \sum t_j /2).$$

2 - There is a lift $SU(n) \to Spin(2n)$ from $SU(n) \to SO(2n)$. Also an embedding (injective group homomorphism) $ SU(n) \subset Spin(2n) $:

Another way to say is this valid fact: "Does the homomorphism $SU(n) \to SO(2n)$ lift to $SU(n) \to Spin(2n)$?"

We can take $T$ to be in $SU(n)$, i.e. take $\prod e^{it_j} =1$. Then $\exp( i \sum t_j /2) = \pm 1$, so $\tilde l (T)$ is actually in $Spin(2n)$.

My questions

The above we had shown $ U(n) \subset Spin^c(2n) = \frac{Spin(2n) \times U(1)}{\mathbf{Z}/2}$. However, when $n=2k+1$, the $Spin^c(2n)$ has a ${\mathbf{Z}/4}$ center. So the $Spin(2n)$ and $U(1)$ can share a common normal subgroup $\mathbf{Z}/4$, more than just a $\mathbf{Z}/2$. I want to prove or disprove the following fact

When $n=2k+1$, is there any valid group homomorphism $$U(n) \to \frac{Spin(2n) \times U(1)}{\mathbf{Z}/4}:=\frac{Spin(4k+2) \times U(1)}{\mathbf{Z}/4} \tag{1}$$ that is also the embedding $ U(n) \subset \frac{Spin(2n) \times U(1)}{\mathbf{Z}/4}?$ Namely, $$ U(2k+1)=\frac{SU(2k+1) \times U(1)}{\mathbf{Z}/(2k+1)} \subset \frac{Spin(4k+2) \times U(1)}{\mathbf{Z}/4}? \tag{2}$$

p.s. If this relation does not hold for general $n=2k+1$, it will be great to know whether certain $n=3,5,7,\dots$, my relations eq.(1) and eq.(2) still hold.

Ref: https://www.maths.ed.ac.uk/~v1ranick/papers/abs.pdf

Answer 1

Let $\omega = e_1e_2\dots e_{2n-1}e_{2n}$.

For $n > 1$, the center of $Spin(2n)$ is $Z(Spin(2n)) = \{\pm 1, \pm\omega\}$. Note that $\omega^2 = (-1)^n$, so

$$Z(Spin(2n)) = \begin{cases} \langle -1, \omega\rangle & n\ \text{is even}\\ \langle\omega\rangle & n\ \text{is odd} \end{cases} \cong \begin{cases} \mathbb{Z}/2\oplus\mathbb{Z}/2 & n\ \text{is even}\\ \mathbb{Z}/4 & n\ \text{is odd.}\end{cases}$$

We also have the central subgroup $\langle i\rangle < U(1)$ which is isomorphic to $\mathbb{Z}/4$, so we can form the quotient of $Spin(2n)\times U(1)$ by the central subgroup $\langle(\omega, i)\rangle\cong\mathbb{Z}/4$. Denote the quotient by $G$.

As $(\omega, i)^2 = (-1, -1)$, there is a natural map $\varphi : Spin^c(2n) \to G$ which has kernel $\langle[(\omega, i)]\rangle$, so the composite map $\varphi\circ\tilde{l} : U(n) \to G$ has kernel $\ker(\varphi\circ\tilde{l}) = \tilde{l}^{-1}(\langle[(\omega, i)]\rangle)$. Since $\tilde{l}$ is an embedding, $\varphi\circ\tilde{l}$ is injective if and only if $[(\omega, i)] \not\in \tilde{l}(U(n))$.

Suppose $\tilde{l}(T) = [(\omega, i)]$. Note that the coefficient of $\omega$ in $\prod_{j=1}^n \left( \cos (t_j/2) + \sin (t_j/2) e_{2j-1}e_{2j} \right)$ is $\prod_{j=1}^n\sin(t_j/2)$ which is $\pm 1$ if and only if $\sin(t_j/2) = \pm 1$ (and hence $\cos(t_j/2) = 0$) for every $j$. It follows that $e^{it_j} = -1$ for all $j$ so $T = -I$. As $\tilde{l}(-I) = [(\omega, i^n)] = [(\omega, (-1)^ki)]$, we see that $\varphi\circ\tilde{l}$ is injective if and only if $k$ is odd.

Example: When $k = 0$, the group $G$ is the quotient of $U(1)\times U(1)$ by $\langle(i, i)\rangle \cong \mathbb{Z}/4$, and the map $U(1) \to G$ is given by $e^{i\theta} \mapsto [(e^{i\theta/2}, e^{i\theta/2})]$ which is not injective because $-1 \mapsto [(i, i)] = [(1, 1)]$.

We also could have taken the quotient of $Spin(2n)\times U(1)$ by the central subgroup $\langle(\omega, -i)\rangle \cong \mathbb{Z}/4$. Arguing as above, the induced map on $U(n)$ is injective if and only if $k$ is odd, as before. Note that the quotients are isomorphic as the map $Spin(2n)\times U(1) \to Spin(2n)\times U(1)$, $(g, z) \mapsto (g, z^{-1})$ descends to an isomorphism.

I don't know if there is an embedding $U(n) \to G$ for $k$ even, but if there is, the diagram

\begin{array}{ccc} & & Spin^c(2n)\\ & \nearrow & \downarrow\\ U(n) & \longrightarrow & G \end{array}

doesn't commute. For $k = 0$, we have $G \cong U(1)\times U(1)$, so there is an embedding $U(1) \to G$.